Is there a chain rule for integration? I know the chain rule for derivatives. The way as I apply it, is to get rid of specific 'bits' of a complex equation in stages, i.e I will derive the $5$th root first in the equation $(2x+3)^5$ and continue with the rest.
I wonder if there is something similar with integration. I tried to integrate that way $(2x+3)^5$ but it doesn't seem to work. Well, it works in the first stage, i.e it's fine to raise in the power of $6$ and divide with $6$ to get rid of the power $5$, but afterwards, if we would apply the chain rule, we should multiply by the integral of $2x+3$!, But it doesn't work like that, we just need to multiply by $1/2$ and that's it.
So my question is, is there chain rule for integrals? I want to be able to calculate integrals of complex equations as easy as I do with chain rule for derivatives.
 A: The chain rule for integration is basically $u$-substitution. 
A: The "chain rule" for integration is the integration by substitution.
$$\int_a^b f(\varphi(t)) \varphi'(t)\text{ d} t  = \int_{\varphi(a)}^{\varphi(b)} f(x) \text{ d} x $$

So in your case we have $f(x) = x^5$ and $\varphi(t) = 2t+3$:
$$
\int (2t + 3)^5 \text{ d}t =
\int {1 \over 2}\left((2t + 3)^5\cdot2\right) \text{ d}t = 
{1\over 2}\int x^5 \text{ d}x = {1\over 12} x^6 + C= {1\over 12} (2t+3)^6 + C$$
A: For calculating derivatives, we use the chain rule by multiplying by one.
$$\frac{dy}{dx}=\frac{dy}{dx}\cdot\frac{du}{du}=\frac{dy}{du}\cdot \frac{du}{dx}$$
Similarly, when integrating with the substitution rule, we also multiply by one. Here is a specific example.
$$\begin{array}{lll}
\displaystyle\int_{x=0}^{x=2}xe^{x^2}dx &=& \displaystyle\int_{x=0}^{x=2}xe^{x^2}\color{red}{dx}\cdot\frac{\frac{dx^2}{\color{red}{dx}}}{\frac{dx^2}{dx}}\\
&=&\displaystyle\int_{x=0}^{x=2}\frac{xe^{x^2}\color{red}{dx}\cdot\frac{dx^2}{\color{red}{dx}}}{\frac{dx^2}{dx}}\\
&=&\displaystyle\int_{x=0}^{x=2}\frac{xe^{x^2}dx^2}{2x}\\
&=&\displaystyle\int_{x=0}^{x=2}\frac{e^{x^2}dx^2}{2}\\
&=&\displaystyle\int_{u=0}^{u=4}\frac{e^{u}du}{2}\\
\end{array}$$
Where $u=x^2$.
Note that the numerator of $\frac{\frac{dx^2}{dx}}{\frac{dx^2}{dx}}$ is interpreted as a ratio of differentials, whereas the denominator is interpreted as a derivative (function).
A: If we know the integral of each of two functions, it does not follow that we can compute the integral of their composite from that information.
Example
$$
\int e^{-x}\;dx = -e^{-x} +C\\
\int x^2\;dx = \frac{x^3}{3} +C\\
$$
but
$$
\int e^{-x^2}\;dx = \frac{\sqrt{\pi}}{2}\;\mathrm{erf}(x) + C
$$
is not an elementary function.
A: It's possible by generalising Faa Di Bruno's formula to fractional derivatives then you can make the order of differentiation negative to obtain a series for for the n'th integral of f(g(x)). 
Here's a paper detailing the fractional chain rule:
Fractional derivative of composite functions: exact results and physical applications,by Gavriil Shchedrin, Nathanael C. Smith, Anastasia Gladkina, Lincoln D. Carr
https://arxiv.org/abs/1803.05018
A: I'm guessing you're asking how to do the integral
$$\int (2x+3)^5 \, dx$$
I would use substitution:
$$u=2x+3 \\
du=2 \, dx$$
So your new integral is
$$\int \frac{u^5}2 \, du = \frac{u^6}{12} +C$$
Then you replace $u$ with the original $2x+3$ to get
$$\int \frac{u^5}2 \, du = \frac{u^6}{12} +C = \frac{(2x+3)^6}{12} +C$$
If you want to see how this relates to the chain rule, take the derivative of your answer, and it should get you the function "inside" the original integral.
$$F(x)=\frac{(2x+3)^6}{12} = f(g(x))$$
$$f(x)=\frac{x^6}{12} \, \, \, g(x)=2x+3 \\
f'(x)=\frac{x^5}2 \, \, \, g'(x)=2 \\$$
Using the chain rule we get
$$F'(x) = f'(g(x))g'(x) = f'(2x+3)g'(x) = \frac{(2x+3)^5}2 (2) = (2x+3)^5$$
A: No.
Sorry for turning up late here, but I think the other (excellent) answers miss a key point. There is no direct, all-powerful equivalent of the differential chain rule in integration. The existence of the chain rule for differentiation is essentially what makes differentiation work for such a wide class of functions, because you can always reduce the complexity. The absence of an equivalent for integration is what makes integration such a world of technique and tricks.
The key point I speak of, therefore, is that hardly any functions can be integrated! Given a function of any complexity, the chances of its antiderivative being an elementary function are very small.
This is deeply contrary to the expectations you build when learning integration - but that's because the lessons are focusing on functions you can integrate, which fortunately overlap closely with the sorts of elementary functions you'd have learned at that stage: trig, exp, polynomials, inverses. They don't focus on the absence of techniques on non-integrable functions, because there's not much to say, and that leaves the impression that having an elementary antiderivative is the norm. And when you think about it, the key technique in integration is spotting how to turn what you've got into the result of a differentiation, so you can run it backwards. 
Fortunately, many of the functions that are integrable are common and useful, so it's by no means a lost battle. And the mine of analytical tricks is pretty deep. And when that runs out, there are approximate and numerical methods - Taylor series, Simpsons Rule and the like, or, as we say nowadays "computers" - for solving anything definite. Or we just give the result a nice name (eg erf) and leave it at that.
Oblig. xkcd: https://xkcd.com/2117/
A: Consider the functions z(y) and y(x).  I am showing an example of a chain rule style formula to calculate
$I(x) = \int dx z(y(x)),$
where z(y) can be triply integrated over dy, and where
$y(x)=\sqrt{x}$ or $y(x)=x.$
To construct a formula for I(x), first define F(y) as the triple integration of z(y) over dy, that is
$F(y) = \iiint dy dy dy z(y).$
Next evaluate F(y) for y(x), that is define
$G(x) = F(y(x)).$
Differentiate G(x) twice over dx and then divide by $(dy/dx)^3,$ yielding
$I(x) = \int dx z(y(x)) = G''(x) / y'^3.$
Consider an example calculation of I(x) where $z = y^3.$
First let $y(x)=\sqrt{x},$ so $dy/dx = (1/2) {x^{-1/2}}.$
Then
$F(y) = y^6 / 120 + ay^2/2 + by + c,$ which yields
$G(x) = F(y(x)) = x^3 /120  + ax/2 + bx^{1/2} + c,$ and then
$G''(x) = x/20 - (1/4)bx^{-3/2},$ so that
$I(x) = y'^{-3} G''(x) = 8 x^{3/2} [ x/20 - (1/4)bx^{-3/2} ]= (2/5)x^{5/2}  - 2b.$
Directly integrating for $y = x^{1/2}$ and $z = y^3$ yields
$I(x) = \int dx z(y(x)) = \int dx y^3  = \int dx x^{3/2} = (2/5) [x^{5/2}] + constant.$
Next for this same example z = $y^3$ let y = x.
Then
$F(y) = y^6/120 + ay^2/2 + by + c,$ which yields
$G(x) = F(y(x)) = x^6/120 + ax^2/2 + bx + c,$  so that
$I(x) = y'^{-3}G''(x)  = 1^{-3} [x^4/4 + a]  = x^4/4 + a.$
Directly integrating yields
$I(x) = \int dx z(y(x)) =  \int dx y^3  =  \int dx x^3   = x^4/4 + constant.$
This demonstrates that the direct and chain rule methods agree with each other to within a constant for $y(x)=x$ and $y(x)=\sqrt{x}$ for the specific function $z(y) = {y^3}.$  This agreement should work for any function z(y) where $y(x)=x$ or $y(x)=\sqrt{x}.$
A: If $g$ is invertible, then the integral
$$
\int f(g(x)) \, dx
$$
can be attacked using integration by substitution. Set $u=g(x)$, meaning that $x=g^{-1}(u)$ and $dx=(g^{-1})'(u)du$. The integral becomes
$$
\int f(u)(g^{-1})'(u) \, du \, .
$$
There are instances where the transformed integral is easier to solve than the original one, but there is no guarantee of this, and obviously we still don't have a 'rule' that tells us that
$$
\int f(g(x)) \, dx = F(x) + C
$$
for some function $F$.

For an example of when this technique might be useful, consider
$$
\int \sqrt{1-x^2} \, dx \, .
$$
This integral looks impossibly hard to solve until you make the substitution $\theta=\arcsin x$. Note that the inner function $g$ is in disguise:
$$
\int \sqrt{1-x^2} \, dx = \int \sqrt{1-\sin^2(\arcsin x)} \, dx \, .
$$
(In practice, we tend to say 'let $x=\sin \theta$' and proceed from there, but this is not quite correct because $x=\sin \theta$ does not imply $\arcsin x =\theta$.)
A: There is no general chain rule for integration known.
The goal of indefinite integration is to get known antiderivatives and/or known integrals.
To get chain rules for integration, one can take differentiation rules that result in derivatives that contain a composition and integrate this rules once or multiple times and rearrange then. For some kinds of integrands, this special chain rules of integration could give known antiderivatives and/or known integrals.
From a chain rule, we expect that the left-hand side of the equation is $\int f(g(x))dx$.
For linear $g(x)$, the commonly known substitution rule
$$\int f(g(x))\cdot g'(x)dx=\int f(t)dt;\ t=g(x)$$
becomes a chain rule.
Further chain rules are written e.g. in
Will, J.: Product rule, quotient rule, reciprocal rule, chain rule and inverse rule for integration. May 2017:
Let
$c$ be an integration constant,
$\gamma$ be the compositional inverse function of function $g$,
$F(g(x))=\int f(t)dt+c;\ t=g(x)$.
$$\int f(g(x))dx=\int f(t)\gamma'(t)dt;\ t=g(x)$$
$$\int f(g(x))dx=xf(g(x))-\int f'(t)\gamma(t)dt;\ t=g(x)$$
$$\int f(g(x))dx=\left(\frac{d}{dx}F(g(x))\right)\int\frac{1}{g'(x)}dx-\int \left(\frac{d^{2}}{dx^{2}}F(g(x))\right)\int\frac{1}{g'(x)}dx\ dx$$
$$\int f(g(x))dx=\frac{F(g(x))}{g'(x)}+\int F(g(x))\frac{g''(x)}{g'(x)^{2}}dx$$
The complexity of the integrands on the right-hand side of the equations suggests that these integration rules will be useful only for comparatively few functions. For linear g(x) however the integrand on the right-hand side of the last equation simplifies advantageously to zero. But this is already the substitution rule above.
